Maths Simple Equation part 7 (Questions 1: Solving Equation ) CBSE Class 7 Mathematics VII Video Lecture | Extra Documents & Videos for Class 7

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well she number 2 so 5m plus 7 is equal
to 70 so here also our two more two is
the same we want to shift all the
constants to the right hand side so here
in the first step what we will do we
will move seven to right hand side so
this side we will have 5m and they said
you will have minus seven to seventeen
minus seven now 5m is equal to 17 minus
7 is 10 now in this side you have 5 into
M so let's move 5 to right hand side so
this becomes 10 by 5 so 5 to 10
therefore M is equal to 2 so this is the
solution so here this side this equation
also let us try to see 3 P divided by 10
is equal to 6 so here first let us try
to move this 10 to right-hand side so
what will happen here it is divided by
suicide it could become multiplied it by
10 so therefore you get 3 P is equal to
60 now this 3 the moment you shift it to
this side it will become divided by you
so 260 divided by 3 so 3 into 20 is 60
so what do you get you get P is equal to
20 as the solution for this equation
question number 3 so on 10 P plus 10 is
equal to hundred here also first let's
move the constants so when you move 10
on the right hand side it becomes 100
minus 10 so you can write it as 10 P is
equal to 90 now when you want to shift
this 10 to right hand side what happens
here it was x so this side
/ 10 so 10 9 and 19 therefore P is equal
to 9 is the solution of this equation
question number 4
Sall 3s plus 12 is equal to 0 so in this
case in the first step we will move this
constant to the side so this becomes 3 s
is equal to 0 minus 12 and what would be
0 minus 12
0 minus 12 would be minus 12 now let's
move 3 to right-hand side
so therefore this becomes minus 12
divided by 3 so 3 4 12 therefore s is
equal to minus 4
Pesci number 5 2 Q minus 6 is equal to 0
similar here also first we will move
this 6 to right-hand side so this
becomes 2 q is equal to 0 plus 6 so we
can write this as 2 q is equal to 6 or
we can say Q is equal to 6 divided by 2
so 2 3 is the 6 therefore Q is equal to
3 is a solution for this equation
question number 6 so on 7m + 19 by 2 is
equal to 13 so in this case we will
first shift the constant term that is 19
by 2 therefore this becomes 7 M is equal
to 13 minus 19 by 2 so let us solve
right-hand side so here we will take LCM
which would be 2 so 1 into 2 so 13 into
2 minus 19 so this is how we had learned
subtraction of fractions right so we
already know this so 7m is equal to 26
minus 19 by 2 all we can say that 7m is
equal to 7 by 2 now here it is 7 into M
so only M would be equal to
in 2/7 so this is nothing but 7 by 2
into 1 by 7 so 7 & 7 will cancel out
therefore M is equal to 1 by 2 is the
solution for this equation question
number 7 3 L by 2 is equal to 2 by 3 so
let us first move this constant term to
the right hand side so what will happen
this is divided by 2 that's why when it
comes to this side it'll become
multiplied by 2 so this implies 3 M is
equal to 4 by 3 or we can say L is equal
to 4 by 3 divided by 3 that is equal to
4 by 3 into 1 by 3 that is equal to 4 by
9 therefore L is equal to 4 by 9 is a
solution for this equation thank you
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